Question Number 132257 by Salman_Abir last updated on 12/Feb/21 Answered by bemath last updated on 12/Feb/21 $$\left(\mathrm{i}\right)\mathrm{E}\left(\mathrm{0},\mathrm{0},\mathrm{8}\right),\:\mathrm{C}\left(\mathrm{3},\mathrm{3},\mathrm{0}\right)\:\mathrm{vector}\: \\ $$$$\:\mathrm{EC}\:=\:\mathrm{3}\hat {\mathrm{i}}+\mathrm{3}\hat {\mathrm{j}}−\mathrm{8}\hat {\mathrm{k}}\:;\:\mid\mathrm{EC}\mid=\sqrt{\mathrm{9}+\mathrm{9}+\mathrm{64}}\:=\sqrt{\mathrm{82}} \\ $$$$\:\mathrm{H}\left(\mathrm{0},\mathrm{3},\mathrm{8}\right)\:,\mathrm{B}\left(\mathrm{3},\mathrm{0},\mathrm{0}\right)\:\mathrm{vector}\: \\…
Question Number 66715 by mr W last updated on 18/Aug/19 $${if}\:{f}\left({x}\right)=\frac{\mathrm{ln}\:\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=? \\ $$ Commented by Tony Lin last updated on…
Question Number 1147 by prakash jain last updated on 04/Jul/15 $${f}\left({f}\left({x}\right)\right)={x}^{\mathrm{2}} −{x}+\mathrm{1} \\ $$$${f}\left(\mathrm{0}\right)=? \\ $$ Commented by prakash jain last updated on 04/Jul/15 $${f}\left({f}\left(\mathrm{0}\right)\right)=\mathrm{1}…
Question Number 132198 by benjo_mathlover last updated on 12/Feb/21 $$\mathrm{If}\:\begin{cases}{\mathrm{16}^{{a}+{b}} \:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\\{\mathrm{16}^{{b}+{c}} \:=\:\mathrm{2}}\\{\mathrm{16}^{{a}+{c}} \:=\:\mathrm{2}\sqrt{\mathrm{2}}}\end{cases} \\ $$$$\:\mathrm{then}\:\mathrm{c}\:=\:\_\_\: \\ $$ Answered by EDWIN88 last updated on 12/Feb/21 $$\mathrm{eq}\left(\mathrm{1}\right)\:\mathrm{16}^{\mathrm{2}\left(\mathrm{a}+\mathrm{b}\right)}…
Question Number 132180 by bounhome last updated on 12/Feb/21 $${solve}\:: \\ $$$$\mathrm{2}{sec}^{\mathrm{2}} {x}+\left(\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{3}\right){secx}−\mathrm{3}\sqrt{\mathrm{2}}=\mathrm{0}\:;\:\mathrm{0}\leqslant{x}\leqslant\frac{\pi}{\mathrm{4}} \\ $$ Answered by benjo_mathlover last updated on 12/Feb/21 $$\:\mathrm{3}\sqrt{\mathrm{2}}\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}−\left(\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{3}\right)\mathrm{cos}\:\mathrm{x}−\mathrm{2}=\mathrm{0} \\…
Question Number 66619 by mr W last updated on 17/Aug/19 $${solve}\:{for}\:{x},{y}\in{R} \\ $$$$\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\mathrm{ln}\:\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}=\frac{\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }}{\mathrm{ln}\:\left({y}+\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\right)} \\ $$ Answered by Smail last updated on…
Question Number 1041 by tera last updated on 22/May/15 $${if}\:{the}\:{roots}\:{of}\:{the}\:{equation}\:{ax}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{2}=\mathrm{0}\:{is}\:\mathrm{2}\:{and}\:{b}. \\ $$$$ \\ $$$${then}\:\mathrm{4}{a}^{\mathrm{2}} −\mathrm{4}{ab}+{b}^{\mathrm{2}} =… \\ $$ Answered by prakash jain last updated…
Question Number 66546 by Sayantan chakraborty last updated on 17/Aug/19 Commented by Sayantan chakraborty last updated on 17/Aug/19 $$\mathrm{urgent}\:\mathrm{need} \\ $$ Commented by ~ À…
Question Number 1007 by tera last updated on 13/May/15 $$\left[{x}−\mathrm{2}\right]>−\mathrm{2}\:.\:\:{x}=….. \\ $$ Answered by prakash jain last updated on 13/May/15 $$\left({x}−\mathrm{2}\right)>−\mathrm{2}\Rightarrow{x}−\mathrm{2}+\mathrm{2}>−\mathrm{2}+\mathrm{2}\Rightarrow{x}>\mathrm{0} \\ $$ Terms of…
Question Number 989 by Madan pd gupta last updated on 13/May/15 $$\mathrm{3}+\mathrm{4} \\ $$ Answered by rpatle69@gmail.com last updated on 13/May/15 $$\mathrm{7} \\ $$ Terms…